题目

若sinx+sin^2 x=1 求cos^2 x+cos^6 x+cos^8 X

提问时间：2020-11-03

答案

sinx+sin^2 x=1
sinx=1-sin^2x=cos^2x
cos^4x=(cos^2x)^2=sin^2x=1-cos^2x=1-sinx
cos^8x=(cos^4x)^2=(1-sinx)^2
=1-2sinx+sin^2x
=1-2sinx+1-cos^2x
=1-2sinx+1-sinx
=2-3sinx
cos^6x=cos^2x*cos^4x=sinx*(1-sinx)
=sinx-sin^2x
=sinx+cos^2x-1
=2sinx-1
cos^2 x+cos^6 x+cos^8 X
=sinx+2sinx-1+(2-3sinx)
=1